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Magnetics question

2021 · 16 Mar · Shift 2 · Q55
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  5. /2021 · 16 Mar · Shift 2 · Q55

Magnetics question

2021 · 16 Mar · Shift 2 · Q55

JEE MainPhysicsMagneticsMCQ+4 / −1
A charge Q is moving dl→\overrightarrow {dl}dl distance in the magnetic field B→\overrightarrow {B}B. Find the value of work done by B→\overrightarrow {B}B.
  1. A
    Zero
  2. B
    −-− 1
  3. C
    Infinite
  4. D
    1
View written solutionFree

Correct answer: A

  1. Magnetic force on a moving charge

For a charge QQQ moving with velocity v⃗\vec vv in a magnetic field B⃗\vec BB, the magnetic force is

F⃗=Q(v⃗×B⃗)\vec F = Q(\vec v \times \vec B)F=Q(v×B)
  1. Direction of magnetic force

The cross product v⃗×B⃗\vec v \times \vec Bv×B is always perpendicular to v⃗\vec vv. So, the magnetic force F⃗\vec FF is always perpendicular to the instantaneous displacement dl⃗d\vec ldl (since displacement is along velocity direction).

  1. Work done by force

The infinitesimal work done is

dW=F⃗⋅dl⃗dW = \vec F \cdot d\vec ldW=F⋅dl

Since F⃗⊥dl⃗\vec F \perp d\vec lF⊥dl,

F⃗⋅dl⃗=0\vec F \cdot d\vec l = 0F⋅dl=0

Hence,

dW=0dW = 0dW=0

and therefore total work done by the magnetic field is also

W=0W = 0W=0
  1. Checking options
  • A: Zero →\rightarrow→ Correct
  • B: −1-1−1 →\rightarrow→ Incorrect
  • C: Infinite →\rightarrow→ Incorrect
  • D: 1 →\rightarrow→ Incorrect
  1. Final answer

The work done by a magnetic field on a moving charge is always zero because magnetic force changes only the direction of motion, not the speed.

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